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Question

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Question about how to know which Force Vectors are positive or negative (or Rx or Ry components?). I know how to setup the problem with all the forces, and to add the y-components or x-components to get each Force on joint , but I'm having an issue how to know whether it's a negative or positive in the equation! Can anyone explain!

For example, FCD is = 50, but in the equation I get FCD -99.8 -249.6(3/5)=0 which gives the wrong answer.

Explanation / Answer

Sign is convention and you follow the convention consistnetly thorugh out the solution.

You may consder the forces acting towards left as -ve then forces towad right will be +ve, similarly forces acting downward are taken -ve and force upward are +ve. when you consider moments of force about a point Clockwise momnets are considered as -ve and Counter clockwise momnets as +ve. This is a general understanding but you can setup your own convetion for signs, you only need to be carefull to follow the same convetion through out the solution. Sign is a way to represent the direction for the force or the moment of the force. One direction you consider as +ve then the opposite direction will be -ve.

First we get the reactions at F and G, for this we ingnore all the members and just look at the forces acting on the structure and the reactions. The only force acting on is at A 100.

summing up the moments about F to be 0.

Rg *4 -100*9

Rg = 225   towards right

summing up the moments about G to be 0

Rfn *4 +100*9

Rfn   = -225   , i.e. left ward

Rfn is the normal reaction at F, vertical reaction Rfv will have 0 moment about G

Consider the forces acting at G

reaction at G and the axial force on meber EG

Feg + Rg = 0

Feg   = -Rg = - 225

Consider the forces acting at A and equate the x and y components

Fab Cos(45) -100 =0 ( Fab upward considerd +ve and 100 downward -ve)

Fab = 141.42

Fac + Fac Cos(45) = 0

Fac = - 100

consder point B and equate the x and y components

Fab Cos(45) + FbdCos(45) =0

Fbd = -Fab = -141.42

Fdb = -Fbd = 141.42 ( this the sign convention, the direction of bd is opposite that of db)

Fdb and F ab are at right angles in triangle ABD

Fbc = - Fcb = -( Fdb*Cos(45) + Fab Cos(45)) = -200

equate forces at C, x and y- components

y-components

Fbc = Fce (4/5)

Fce = -200*5/4 = -250

x-components

- Fcd + Fce(3/5) = Fca = -Fac = 100

Fcd = -100 + 150 = 50

equate x and y components at D

Fdb * cos(45) = Fde

Fde = 141.42 *Cos(45) = 100

Fdf = FdbCos(45) + Fcd = 100 + 50 = 150 ( The force in member CD is 50 at C it is towards D and at D it is towards C, equal and oppoiste reaction, Newton's third law)

equating the x-components at E

Fef (3/5) + Feg = Fce(3/5)

Fef = (225 - 250*3/5 )*5/3 = 125

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